\(\frac{x+y}{7x+y}-\frac{6x}{-7x}=\frac{x+y}{7x+y}+\frac{6x}{7x}\)

\(=\frac{49x+13y}{7\left(7x+y\right)}=\frac{49x+13y}{49x+7y}\)

\(=1+\frac{6y}{49x+7y}\)

A = (2x - 3)(x² + 4x) - 2(x³ + 2x + 6)

   = 2x(x² + 4x) - 3(x² + 4x) - 2x³ - 4x - 12

  = 2x³ + 8x² - 3x² - 12x - 2x³ - 4x - 12

  = 5x² - 16x - 12

Thay x = 4 vào biểu thức trên ta có : 5.4² - 16.4 - 12 = 4

B = x(x² + 7x) - (x + 9)(x² + 17)

   = x³ + 7x² - x(x² + 17) - 9(x² + 17)

  = x³ + 7x² - x³ - 17x - 9x² - 153

  = -2x² - 17x - 153 

Thay x = 5 vào biểu thức trên ta có : -2.5² - 17.5  - 153 = -50 - 85 - 153 = -288

A = ( 2x - 3 )( x² + 4x ) - 2( x³ + 2x + 6 )

A = 2x³ + 8x² - 3x² - 12x - 2x³ - 4x - 12

A = 5x² - 16x - 12

Thế A = 4 ta được :

A = 5.4² - 16.4 - 12 = 4

B = x( x² + 7x ) - ( x + 9 )( x² + 17 )

B = x³ + 7x² - ( x³ + 17x + 9x² + 153 )

B = x³ + 7x² - x³ - 17x - 9x² - 153

B = -2x² - 17x - 153

Thế x = 5 ta được :

B = -2.5² - 17.5 - 153 = -288

Dài quá ! Nên vẫn phải làm ^_^.

Bài 1: 

+) \(A=x^2-2x+6=x^2-2x+1+5=\left(x-1\right)^2+5\ge5\)

Min A = 5 \(\Leftrightarrow x=1\)

+) \(B=x^2+6x+12=x^2+6x+9+3=\left(x+3\right)^2+3\ge3\)

Min B = 3 \(\Leftrightarrow x=-3\)

+) \(C=4-x^2+2x=-\left(x^2-2x+4\right)=-\left[\left(x-1\right)^2+3\right]=-\left(x-1\right)^2-3\le-3\)

Max C = -3 \(\Leftrightarrow x=1\)

+) \(D=-x^2+6x=-\left(x^2-6x+9-9\right)=-\left(x-3\right)^2+9\le9\)

Max D = 9 \(\Leftrightarrow x=3\)

Bài 2 :

a) \(x^2-x-3x+3=0\)

\(\Leftrightarrow x^2-4x+4-1=0\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

b) \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

c) Xem lại đề hộ mình nha 

d) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)

\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)

\(=-xy^2+yx^2-yz^2+zy^2-xz^2+zx^2\)

\(=xy^2\left(1-1\right)+yz^2\left(1-1\right)+zx^2\left(1-1\right)\)

\(=\left(xy^2+yz^2+zx^2\right).0\left(=0\right)\)

Ta có :

\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)

\(=\left(x+y\right)^2.\left(x-y\right)+\left(y+z\right).\left(y^2-x^2+x^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)

\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2+z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)

\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2\right)-\left(y+z\right)\left(z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)

\(=\left(x^2-y^2\right)\left(x+y-y-z\right)-\left(z^2-x^2\right).\left(y+z-z-x\right)\)

\(=\left(x^2-y^2\right).\left(x-z\right)-\left(z^2-x^2\right).\left(y-x\right)\)

\(=\left(x-y\right)\left(x+y\right)\left(x-z\right)+\left(z-x\right)\left(z+x\right)\left(x-y\right)\)

\(=\left(x-y\right).\left[\left(x+y\right)\left(x-z\right)+\left(z-x\right).\left(x+z\right)\right]\)

\(=\left(x-y\right)\left(x^2-zx+xy-yz+zx+z^2-x^2-xz\right)\)

\(=\left(x-y\right)\left(z^2-zx+xy-yz\right)\)

\(=\left(x-y\right)\left[z.\left(z-x\right)-y.\left(z-x\right)\right]\)

\(=\left(x-y\right)\left(z-y\right)\left(z-x\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

Ta có :

\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)

\(=\left(x+y\right)^2.\left(x-y\right)+\left(y+z\right).\left(y^2-x^2+x^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)

\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2+z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)

\(=\left(x+y\right)\left(x^2-y^2\right)-\left(y+z\right)\left(x^2-y^2\right)-\left(y+z\right)\left(z^2-x^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)

\(=\left(x^2-y^2\right)\left(x+y-y-z\right)-\left(z^2-x^2\right).\left(y+z-z-x\right)\)

\(=\left(x^2-y^2\right).\left(x-z\right)-\left(z^2-x^2\right).\left(y-x\right)\)

\(=\left(x-y\right)\left(x+y\right)\left(x-z\right)+\left(z-x\right)\left(z+x\right)\left(x-y\right)\)

\(=\left(x-y\right).\left[\left(x+y\right)\left(x-z\right)+\left(z-x\right).\left(x+z\right)\right]\)

\(=\left(x-y\right)\left(x^2-zx+xy-yz+zx+z^2-x^2-xz\right)\)

\(=\left(x-y\right)\left(z^2-zx+xy-yz\right)\)

\(=\left(x-y\right)\left[z.\left(z-x\right)-y.\left(z-x\right)\right]\)

\(=\left(x-y\right)\left(z-y\right)\left(z-x\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

a) \(\frac{1}{x+2}+\frac{2}{x+3}=\frac{6}{x+4}\)

ĐKXĐ \(x\ne-2,-3,-4\)

=> \(\frac{1}{x+2}+\frac{2}{x+3}-\frac{6}{x+4}=0\)

=> \(\frac{3x+7}{\left(x+2\right)\left(x+3\right)}-\frac{6}{x+4}=0\)

=> \(\frac{\left(3x+7\right)\left(x+4\right)-6\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+4\right)}=0\)

=> (3x + 7)(x + 4) - 6(x² + 5x + 6) = 0

=> 3x² + 19x + 28 - 6x² - 30x - 36 = 0

=> -3x² - 11x - 8 = 0

=> -3x² - 3x - 8x - 8 = 0

=> -3x(x + 1) - 8(x + 1) = 0

=> (x + 1)(-3x - 8) = 0

=> \(\orbr{\begin{cases}x=-1\\x=-\frac{8}{3}\end{cases}}\)

Vậy ...

b) Thiếu dữ liệu cuả đề 

c) \(\frac{6x+22}{x+2}-\frac{2x+7}{x+3}=\frac{x+4}{x^2+5x+6}\)

ĐKXĐ \(x\ne-2;-3\)

=> \(\frac{\left(6x+22\right)\left(x+3\right)-\left(x+2\right)\left(2x+7\right)}{\left(x+2\right)\left(x+3\right)}=\frac{x+4}{\left(x+2\right)\left(x+3\right)}\)

=> \(6x^2+40x+66-x\left(2x+7\right)-2\left(2x+7\right)=x+4\)

=> \(6x^2+40x+66-2x^2-7x-4x-14=x+4\)

=> 4x² + 29x + 52 = x + 4

=> 4x² + 29x + 52 - x - 4 = 0

=> 4x² + 28x + 48 = 0

=> 4(x² + 7x + 12) = 0

=> x² + 7x +12 = 0

=> x² + 3x + 4x + 12 = 0

=> x(x + 3) + 4(x + 3) = 0

=> (x + 3)(x + 4) = 0

=> \(\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\) 

Mà \(x\ne-2,-3\)nên x = -3 loại

Vậy x = -4

a) Với giá trị của x thì phân thức được xác định là : \(x^2-1\ne0\)

=> \(x^2\ne\pm1\)

b) Rút gọn A : \(A=\frac{x^2+2x+1}{x^2-1}=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{x+1}{x-1}\)

c) Tại x = -2 thì \(A=\frac{\left(-2\right)+1}{\left(-2\right)-1}=\frac{-1}{-3}=\frac{1}{3}\)

d) Ta có : \(A=\frac{x^2+2x+1}{x^2-1}=\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}\)

=> \(2⋮x-1\)=> x - 1 \(\in\)Ư(2) = { \(\pm1;\pm2\)}

+) x - 1 = 1 => x = 2 ; x - 1 = -1 => x = 0

+) x - 1 = 2 => x = 3 ; x - 1 = -2 => x = -1

Vậy : ....

a) Phân thức xác định 

\(\Leftrightarrow x^2-1\ne0\)

\(\Leftrightarrow x\ne\pm1\)

Vậy với \(x\ne\pm1\)thì giá trị của phân thức đã cho xác định.

b) \(A=\frac{x^2+2x+1}{x^2-1}\)

\(\Leftrightarrow A=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow A=\frac{x+1}{x-1}\)

c) x = -2 ( thỏa mãn đkxđ )

Vậy \(A=\frac{-2+1}{-2-1}=\frac{-1}{-3}=\frac{1}{3}\)

d)  A có giá trị nguyên 

\(\Leftrightarrow\frac{x+1}{x-1}\)có giá trị nguyên 

\(\Leftrightarrow\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}\)có giá trị nguyên

\(\Leftrightarrow x-1\inƯ\left(2\right)\)

\(\Leftrightarrow x=\left\{2;3;0\right\}\)

Ta có: ( x - y) z³ + ( y - z ) x³ + ( z - x ) y³  

= ( x - y ) z³ + ( y - z )x³ + ( z - y)y³ + ( y - x ) y³

= ( x - y ) ( z³ - y³ ) + ( y - z ) ( x³ - y³) 

= ( x - y ) ( z - y ) ( z² + zy + y² ) + ( y - z ) ( x - y) ( x² + xy + y² ) 

= ( x - y ) ( y - z ) ( x² + xy + y² - z² - zy - y²) 

= ( x - y ) ( y - z ) [ ( x² - z²) + ( xy - zy) ]

= ( x - y ) ( y - z ) [ ( x - z ) ( x + z ) + y ( x - z ) ]

= ( x - y ) ( y - z ) ( x - z ) ( x + y + z )